If the roots of \(x^2+px+q=0\) are \(p+2\) and \(q-2\), and \(p+q=8\), what is the value of \(p\)?
Answer and explanation
Correct answer: -8
By Vieta's relations for \(x^2+px+q=0\), the sum of roots equals \(-p\). The given roots sum to \((p+2)+(q-2)=p+q=8\), so \(-p=8\) and hence \(p=-8\). Note: checking the product condition \((p+2)(q-2)=q\) with \(p=-8\) (so \(q=16\)) gives \(-84\neq16\), so the full set of data is inconsistent; nevertheless the value of \(p\) is determined by the sum relation asked in the question. Exam tip: check both sum and product; a common mistake is to miss the negative sign and pick \(8\).
Frequently asked questions
What is the correct answer to this question?
-8
Why is this the correct answer?
By Vieta's relations for \(x^2+px+q=0\), the sum of roots equals \(-p\). The given roots sum to \((p+2)+(q-2)=p+q=8\), so \(-p=8\) and hence \(p=-8\). Note: checking the product condition \((p+2)(q-2)=q\) with \(p=-8\) (so \(q=16\)) gives \(-84\neq16\), so the full set of data is inconsistent; nevertheless the value of \(p\) is determined by the sum relation asked in the question. Exam tip: check both sum and product; a common mistake is to miss the negative sign and pick \(8\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
